Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given

1->2->3->4->5->NULL

, m = 2 and n =
4,

return

1->4->3->2->5->NULL

.

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.

</pre><pre name="code" class="java">/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if(m==n){
return head;
}
ListNode dummy = new ListNode(0);
dummy.next=head;
ListNode preM= dummy;
ListNode nodeM=dummy;

//find preM, M
for(int i=0;i<m;i++){
preM=nodeM;
nodeM=nodeM.next;
}
ListNode bfcur=nodeM;
ListNode cur=nodeM.next;
ListNode aftcur;
for(int i=m+1;i<=n;i++){
aftcur=cur.next;
cur.next=bfcur;
bfcur=cur;
cur=aftcur;

}
preM.next=bfcur;
nodeM.next=cur;
return dummy.next;
}

}

看到的另外一种解法是用arraylist(我觉得用stack 更好)来记录m到n的每一个节点的val， 然后把m到n 这些节点的值 reverse， 这也是可以的，但我觉得这道题的目的还是为了考察linked list 结点的 reverse 吧。